8o6 POPULAR SCIENCE MONTHLY. 



demonstrations tliat lie does not understand. Taking the parallelo- 

 gram, for example, let us suppose a figure made like Fig. 4, and we 

 saw through it along the lines A A' and B C. It does not need a 

 very great effort of attention to recognize, experimentally if need 

 be, that the two triangles A A' D and B B' C may be placed one 

 upon the other and are identical. If, from the figure thus formed, 

 we take away the right-hand triangle the parallelogram will remain ; 

 if we take away the other triangle a rectangle Avill be left, or a 

 peculiar parallelogram, of which also we give the idea to the child 

 as a figure in which the angles are formed by straight lines per- 

 pendicular to one another. Here, then, the child gains the notion 

 of the equivalence of a parallelogram and a rectangle of the same 

 base and height ; and this notion, obtained by cutting up a piece of 

 board or pasteboard, he will carry so seriously and firmly in his head 

 that he will never lose it. By cutting the same parallelogram in 

 two, along a diagonal A C, it may be easily shown that the two tri- 

 angles can be placed exactly one upon the other, and that, conse- 

 quently, they have equal areas. These lessons constitute a series of 

 classical theorems in geometry which the child can try with his fin- 

 gers and learn without even giving them the form of theorems. I 

 might show the same as to the area of the trapeze and with many 

 other theorems, but my purpose is only to present as many examples 

 as will make my idea understood, without going into details. 



Yet I can not leave this subject without showing how we can 

 make a very child understand some of the geometrical theorems 



that have acquired a bad repu- 



tation in the world of candi- 

 dates for degrees, including 

 even such as the pons asino- 

 rum of Pythagoras; the dem- 

 onstration, that is, that if we 

 fia 4 construct the triangles B and 



C on the sides of a right-an- 

 gled triangle, their sum will be equal to the square A constructed 

 on the hypotenuse. The usual demonstration of this theorem is 

 not very complicated, but there is something tiresome, artificial, 

 and hard in it. The demonstration I propose is almost intuitive, 

 and the reasoning of it is both simple and rigorous. 



Suppose we take two equal squares, and, making equal lengths 

 on the four sides of one of them, join the points so obtained as in- 

 dicated in the first of the two figures (Figs. 5 and 6) so as to form 

 four right-angled triangles, and then place four other squares in the 

 corners of the original square. These right-angled triangles are of 

 such sort that the sum of their sides is equal to the side of the square. 



